Question: A right circular cylinder with radius 2 is inscribed in a hemisphere with radius 5 so that its bases are parallel to the base of the hemisphere.  What is the height of this cylinder?
Explanation: We draw and label a diagram as follows: [asy]

size(110);
pair O = (0,0); pair A = (.3,.94); pair B = (.3,.075);
draw(O--A--B--cycle,heavycyan);
label("$O$",O,W); label("$A$",A,N); label("$B$",B,S);
import solids; import three; defaultpen(linewidth(0.8)); currentprojection = orthographic(5,0,1.3);
revolution c = cylinder((0,0,0), .4, .91);
draw(c,black);

draw(scale(1,.25)*arc((0,0),1,0,180),dashed);
draw(scale(1,.25)*arc((0,0),1,180,360));
draw(Arc((0,0),1,0,180));

[/asy]

Let the center of the hemisphere be $O$, and let $A$ be a point on the circumference of the top circle of the cylinder.  Since the cylinder is inscribed in the hemisphere, $A$ lies on the hemisphere as well, so $OA=5$.  We drop a perpendicular from $A$ to the base of the hemisphere and let it intersect the base of the hemisphere at $B$.  Since the cylinder is right and $AB$ is a height of the cylinder, $\angle OBA$ is a right angle, and $B$ lies on the circumference of the bottom circle of the cylinder.  Thus, $OB$ is a radius of the cylinder, so $OB=2$. We have that $\triangle OBA$ is right, so by the Pythagorean theorem, we have \[AB=\sqrt{OA^2-OB^2}=\sqrt{5^2-2^2}=\sqrt{21}.\]Thus, the height of the cylinder is $\boxed{\sqrt{21}}$.